The equation of one of the common tangents of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of a tangent to the parabola $y^2=x$ with slope $m$ is given by $y = mx + \frac{1}{4m}$.For this to be a tangent to the circle $x^2+y

Vedclass · Accepted Answer

$x-2y+1=0$

Vedclass · Answer

(D) The equation of a tangent to the parabola $y^2=x$ with slope $m$ is given by $y = mx + \frac{1}{4m}$.For this to be a tangent to the circle $x^2+y^2-6y+4=0$,the perpendicular distance from the center $(0, 3)$ to the line $mx - y + \frac{1}{4m} = 0$ must equal the radius of the circle.The circle equation can be written as $x^2 + (y-3)^2 = 5$,so the radius is $\sqrt{5}$.Thus,$\frac{|m(0) - 3 + \frac{1}{4m}|}{\sqrt{m^2 + 1}} = \sqrt{5}$.Squaring both sides: $\frac{(-3 + \frac{1}{4m})^2}{m^2 + 1} = 5$.$9 - \frac{3}{2m} + \frac{1}{16m^2} = 5m^2 + 5$.Multiplying by $16m^2$: $144m^2 - 24m + 1 = 80m^4 + 80m^2$.$80m^4 - 64m^2 + 24m - 1 = 0$.Testing $m = 1/2$: $80(1/16) - 64(1/4) + 24(1/2) - 1 = 5 - 16 + 12 - 1 = 0$.So,$m = 1/2$ is a solution.The equation of the tangent is $y = \frac{1}{2}x + \frac{1}{4(1/2)} = \frac{1}{2}x + \frac{1}{2}$.Multiplying by $2$: $2y = x + 1$,which simplifies to $x - 2y + 1 = 0$.

The equation of one of the common tangents of the circle $x^2+y^2-6y+4=0$ and the parabola $y^2=x$ is

Similar Questions

Two tangents to the circle $x^2+y^2=4$ at the points $A$ and $B$ meet at $P(-4,0)$. Then the area of quadrilateral $PAOB$,where $O$ is the origin,is

The point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ is

The line $y = mx + c$ will be a normal to the circle with radius $r$ and centre at $(a, b)$,if

If two tangents are drawn from the point $P$ on the circle $x^2+y^2=4$ to the circle $x^2+y^2=1$,where the point $P$ is given by $(\sqrt{2}, \sqrt{2})$,then the slopes of the tangents are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module